Solved papers for JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

done JEE Main Online Paper (Held On 23 April 2013) Total Questions - 1

• question_answer1)                 If\[\operatorname{S}={{\tan }^{-1}}\left( \frac{1}{{{\operatorname{n}}^{2}}+\operatorname{n}+1} \right)+\]                 \[{{\tan }^{-1}}\left( \frac{1}{{{\operatorname{n}}^{2}}+3\operatorname{n}+3} \right)+.......\]                 \[+{{\tan }^{-1}}\left( \frac{1}{1+(\operatorname{n}+19)(\operatorname{n}+20)} \right),\]then tan S is equal to :     JEE Main Online Paper ( Held On 23  April 2013 )
  A)
                   \[\frac{20}{401+20\operatorname{n}}\]                              
  done clear
  B)
                                          \[\frac{\operatorname{n}}{{{\operatorname{n}}^{2}}+20\operatorname{n}+1}\]
  done clear
  C)
                                          \[\frac{20}{{{\operatorname{n}}^{2}}+20\operatorname{n}+1}\]                           
  done clear
  D)
                                          \[\frac{\operatorname{n}}{401+20\operatorname{n}}\]
  done clear
  View Answer play_arrow